NIMCET 2027 Progressions Previous Year Questions (PYQs) – Page 3 of 5
Study Stuff for MCA Examinations
Solution
Sum to infinity: $S_{\infty} = \dfrac{a}{1-r}$
Sum of first two terms: $S_2 = a + ar = a(1+r)$
Step 2: Set up the equation
Given: $S_{\infty} = 2 \times S_2$
$\dfrac{a}{1-r} = 2 \cdot a(1+r)$
Step 3: Simplify
Dividing both sides by $a$ (since $a \ne 0$):
$\dfrac{1}{1-r} = 2(1+r)$
$1 = 2(1+r)(1-r)$
$1 = 2(1 - r^2)$
$1 = 2 - 2r^2$
$2r^2 = 1$
$r^2 = \dfrac{1}{2}$
$r = \pm\dfrac{1}{\sqrt{2}}$
Solution
Given: The sequence is in H.P. with \(a_m = n\) and \(a_n = m\) \((m \ne n)\).
In an H.P., reciprocals of terms form an A.P. So, \(\tfrac{1}{a_m} = \tfrac{1}{n}\) and \(\tfrac{1}{a_n} = \tfrac{1}{m}\).
This A.P. leads to common difference \(d = \tfrac{1}{mn}\) and first term \(A = \tfrac{1}{mn}\).
Thus, the reciprocal of the \((m+n)\)th term is:
\(b_{m+n} = \tfrac{m+n}{mn}\)
Hence, the \((m+n)\)th term of H.P. is:
\(a_{m+n} = \dfrac{mn}{m+n}\)
Solution
Short Solution:
Let the two numbers be \( a \) and \( b \).Given: Harmonic mean is 4: $$\frac{2ab}{a + b} = 4 \quad \text{(1)}$$ Arithmetic mean \( A = \frac{a + b}{2} \),
Geometric mean \( G = \sqrt{ab} \)
Given: $$2A + G^2 = 27$$ $$2 \cdot \frac{a + b}{2} + ab = 27 \Rightarrow a + b + ab = 27 \quad \text{(2)}$$ From (1): Multiply both sides by \( a + b \): $$2ab = 4(a + b) \Rightarrow ab = 2(a + b) \quad \text{(3)}$$ Substitute (3) into (2): $$a + b + 2(a + b) = 27 \Rightarrow 3(a + b) = 27 \Rightarrow a + b = 9$$ Then from (3): $$ab = 2 \cdot 9 = 18$$ Now solve: $$x^2 - (a + b)x + ab = 0 \Rightarrow x^2 - 9x + 18 = 0$$ $$\Rightarrow x = 3, 6$$
Final Answer:
$$\boxed{3 \text{ and } 6}$$
Solution
Solution
Solution:
Even numbers from 100 to 998: count $= \frac{998-100}{2}+1=450$.
Exclude evens divisible by $3$ or $5$ using inclusion–exclusion:
- Multiples of $6$ in $[100,998]$: $\lfloor 998/6 \rfloor-\lfloor 99/6 \rfloor = 166-16=150$.
- Multiples of $10$ in $[100,998]$: $\lfloor 998/10 \rfloor-\lfloor 99/10 \rfloor = 99-9=90$.
- Multiples of $30$ in $[100,998]$: $\lfloor 998/30 \rfloor-\lfloor 99/30 \rfloor = 33-3=30$.
Forbidden $=150+90-30=210$ ⇒ Allowed $=450-210={240}$.
Solution
Let the first term of the arithmetic progression be $a$.
Common difference is:
$d=\frac{3}{2}$
Sum of first $11$ terms is:
$S_{11}=88$
Formula for sum of first $n$ terms:
$S_n=\frac{n}{2}[2a+(n-1)d]$
So,
$88=\frac{11}{2}[2a+10d]$
Substitute $d=\frac{3}{2}$:
$88=\frac{11}{2}\left[2a+10\times \frac{3}{2}\right]$
$88=\frac{11}{2}[2a+15]$
Now,
$2a+15=16$
$2a=1$
$a=\frac{1}{2}$
The $10^{\text{th}}$ term is:
$T_{10}=a+9d$
$T_{10}=\frac{1}{2}+9\times \frac{3}{2}$
$T_{10}=\frac{1}{2}+\frac{27}{2}$
$T_{10}=14$
The $11^{\text{th}}$ term is:
$T_{11}=a+10d$
$T_{11}=\frac{1}{2}+10\times \frac{3}{2}$
$T_{11}=\frac{1}{2}+15$
$T_{11}=\frac{31}{2}$
So, the roots of $3x^2-px+q=0$ are $14$ and $\frac{31}{2}$.
For equation $3x^2-px+q=0$,
Sum of roots:
$\frac{p}{3}=14+\frac{31}{2}$
$\frac{p}{3}=\frac{28+31}{2}$
$\frac{p}{3}=\frac{59}{2}$
$p=\frac{177}{2}$
Product of roots:
$\frac{q}{3}=14\times \frac{31}{2}$
$\frac{q}{3}=217$
$q=651$
Now,
$q-2p=651-2\times \frac{177}{2}$
$q-2p=651-177$
$q-2p=474$
Therefore, the correct answer is option $2$.Given,
Solution
Solution
Given equation is
$\left(\sum_{k=1}^{n}(-1)^{k-1}k\right)^2-\sum_{k=1}^{n}(-1)^{k-1}k^2+2450=0$
Check for odd $n$.
Let
$n=2m-1$
Then,
$\sum_{k=1}^{n}(-1)^{k-1}k=1-2+3-4+\cdots+(2m-1)$
Pairing terms,
$(1-2)+(3-4)+\cdots+[(2m-3)-(2m-2)]+(2m-1)$
$=-(m-1)+(2m-1)$
$=m$
So,
$\left(\sum_{k=1}^{n}(-1)^{k-1}k\right)^2=m^2$
Now,
$\sum_{k=1}^{n}(-1)^{k-1}k^2=1^2-2^2+3^2-4^2+\cdots+(2m-1)^2$
For odd $n=2m-1$,
$\sum_{k=1}^{n}(-1)^{k-1}k^2=m(2m-1)$
Now put in the equation:
$m^2-m(2m-1)+2450=0$
$m^2-2m^2+m+2450=0$
$-m^2+m+2450=0$
$m^2-m-2450=0$
Now factorize:
$m^2-m-2450=0$
$m^2-50m+49m-2450=0$
$m(m-50)+49(m-50)=0$
$(m-50)(m+49)=0$
Since $m$ is positive,
$m=50$
Therefore,
$n=2m-1$
$n=2(50)-1$
$n=99$
Solution
Since $a, H_1, H_2, \ldots, H_n, b$ is in HP,
$\dfrac{1}{a},\ \dfrac{1}{H_1},\ \ldots,\ \dfrac{1}{H_n},\ \dfrac{1}{b}$ is in AP with $n+2$ terms.
Step 2: Find common difference $d$
$d = \dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{n+1} = \dfrac{a-b}{ab(n+1)}$
Step 3: Find $H_1$ and $H_n$
$\dfrac{1}{H_1} = \dfrac{1}{a} + d = \dfrac{a+bn}{ab(n+1)}$ $\Rightarrow H_1 = \dfrac{ab(n+1)}{a+bn}$
$\dfrac{1}{H_n} = \dfrac{1}{b} - d = \dfrac{an+b}{ab(n+1)}$ $\Rightarrow H_n = \dfrac{ab(n+1)}{an+b}$
Step 4: Evaluate $\dfrac{H_1+a}{H_1-a}$ using Componendo-Dividendo
$\dfrac{H_1}{a} = \dfrac{b(n+1)}{a+bn}$
Applying componendo-dividendo:
$\dfrac{H_1+a}{H_1-a} = \dfrac{b(n+1)+(a+bn)}{b(n+1)-(a+bn)} $
Step 5: Evaluate $\dfrac{H_n+b}{H_n-b}$ using Componendo-Dividendo
$\dfrac{H_n}{b} = \dfrac{a(n+1)}{an+b}$
Applying componendo-dividendo:
$\dfrac{H_n+b}{H_n-b} = \dfrac{a(n+1)+(an+b)}{a(n+1)-(an+b)} $
Step 6: Add (1) and (2)
$\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} $
$= \dfrac{a+b(2n+1) - b - a(2n+1)}{b-a}$
$= \dfrac{(a-b) + (2n+1)(b-a)}{b-a}$
$= \dfrac{(b-a)(2n+1)-(b-a)}{b-a}$
$= (2n+1) - 1$
$= 2n$
Answer: $\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} = \boxed{2n}$
Solution
$\dfrac{a_1 + a_2 + \cdots + a_n}{n} \geq (a_1 \cdot a_2 \cdots a_n)^{1/n}$
Step 1: Rewrite the sum
$a_1 + a_2 + \cdots + a_{n-1} + 2a_n$
This has $n$ terms: $(a_1, a_2, \ldots, a_{n-1}, 2a_n)$
Step 2: Apply AM-GM
$\dfrac{a_1 + a_2 + \cdots + a_{n-1} + 2a_n}{n} \geq (a_1 \cdot a_2 \cdots a_{n-1} \cdot 2a_n)^{1/n}$
$\geq (2 \cdot a_1 a_2 \cdots a_n)^{1/n}$
$\geq (2c)^{1/n}$
Step 3: Find minimum
$a_1 + a_2 + \cdots + 2a_n \geq n(2c)^{1/n}$
Minimum value $= n(2c)^{1/n}$
Equality holds when $a_1 = a_2 = \cdots = a_{n-1} = 2a_n$
Answer: Minimum value $= \boxed{n(2c)^{1/n}}$
NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
